Probabilistic Cellular Automaton and Markov Chains Matthias Schulz - - PowerPoint PPT Presentation

Probabilistic Cellular Automaton and Markov Chains Matthias Schulz University of Karlsruhe 1

Contents:

• Introduction
• PCAs with 2 states, M1-Neighbourhood
• PCAs with k states, M1-Neighbourhood
• PCAs with 2 states, Mm-Neighbourhood
• Conclusion

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Probabilistic Cellular Automaton:

• Space R, e.g. Zn (Rings)
• Neighbourhood N, e.g. (-1, 0, 1)
• Set of states S, e.g. {0, 1}
• Local transition rule δ, e.g.

δ(012)(0) = 0.4 δ(012)(1) = 0.2 δ(012)(2) = 0.4 In each step, for every cell p in a configuration c = (c1, ..., cn) the probability to go into state i is δ(c(p + N))(i). 3

Markov Chain:

• Set of states S
• Matrix P of transition probabilities:

Pij contains the probability for state i changing to state j. Each Markov Chain induces a directed graph whose nodes are the states; there exists an edge beginning in i and ending in j, iff Pij > 0. 4

Example: The matrix   0 0.5 0.5 0 0 1 0 1   induces the graph 5

A subset M in S is closed, iff Pij = 0 for i ∈ M, j / ∈ M. A closed subset M in S is irreducible, iff there is no nontrivial closed subset in M: Here the set {B, C} is closed and irreducible. 6

Let Mv ⊆ N be the set of all lengths of paths beginning and ending in v ∈ S. v is aperiodic, iff gcd(Mv) = 1: There is the path ABCA of length 3 beginning and ending in A, and the path ACA of length 2 beginning and ending in A; since the gcd of 3 and 2 is 1, A is aperiodic. 7

A Markov Chain is irreducible, iff S is irreducible: 8

A Markov Chain is aperiodic, iff each state in S is aperiodic: 9

A Markov Chain is ergodic, iff it is irreducible and aperiodic: 10

Observation: PCAs can be interpreted as Markov Chains. Question: How hard is it determining whether a given PCA is ergodic? 11

• Introduction
• PCAs with 2 states, M1-Neighbourhood
• PCAs with k states, M1-Neighbourhood
• PCAs with 2 states, Mm-Neighbourhood
• Conclusion

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Definitions: A string (c1, ..., ck) ∈ {0, 1, X}k is a structure, iff

• none of the ci equals X or
• k ≥ 3 and not all ci are X.

A configuration d is an instance of a structure c, if it differs from c only where ci = X, e.g. 101 and 100 are instances of 10X. 13

A transition rule δ is homogenous, iff δ(000)(1) = 1 and δ(111)(0) = 1. A transition rule δ is structure-invariant, iff

• δ is not homogenous and
• there exists a structure d = (d1, ..., dk), such that each successor of an instance of d is an

instance of d (possibly shifted),

• eg. for each δ satisfying δ(101)(1) = 1, δ(010)(0) = 1, the structure 10 is left invariant,

since 10 has 01 as successor, which is 10 shifted to the right. 14

A transition rule δ is inhomogenous, iff

• δ is neither homogenous nor structure-invariant, and
• for odd n the set of inhomogenous configurations is closed.

A transition rule is ergodic, iff each PCA with this transition rule and n ≥ 4 is ergodic. 15

Theorem 1: A transition rule is ergodic, iff it is neither homogenous, structure-invariant nor inhomogenous. This can be shown by analysis of 21 different cases. 16

• Introduction
• PCAs with 2 states, M1-Neighbourhood
• PCAs with k states, M1-Neighbourhood
• PCAs with 2 states, Mm-Neighbourhood
• Conclusion

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Theorem 2: The problem to decide whether or not a given PCA with k states is ergodic is PSPACE-complete in (n + k). Theorem 3: The problem to determine for a given transition rule δ whether or not there exists n ∈ N so that the PCA defined by δ and n is ergodic is undecidable. 18

Proof of Theorem 2: The problem is in PSPACE: Irreducibility and aperiodicity can be checked nondeterminitically with algorithms needing polynomial space in (n + k). With Savitch’s Theorem, it follows that there exists a deterministic algorithm needing poly- nomial space in (n + k). 19

The problem is PSPACE-hard: There is a nondeterministic LBA so that it is PSPACE-hard in n to decide whether or not a given word of length n can be accepted by the LBA. Idea: For each n there is a PCA P constructed with O(n) states and a length in O(n), such that P is ergodic ⇔ a given word w of length n is accepted by a fixed LBA A. 20

Construction:

• 0. The length of P shall be odd and ≥ n + 2.

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• 1. There is a state X which can change into each state and can spread over the PCA;

all other states shall consist of several registers: ... X ... 22

• 1. There is a state X which can change into each state and can spread over the PCA;

all other states shall consist of several registers: ... XXX ... 23

• 1. There is a state X which can change into each state and can spread over the PCA;

all other states shall consist of several registers: ... XXXXX ... It follows that P is ergodic iff from every possible configuration there is a configuration containing X reachable. 24

• 2. Exactly one cell is marked with START, or there is a configuration reachable containing

X (this can be achieved for any odd length): START ... So if there are configurations from which no configurations containing X can be reached, these contain exactly one START-cell. 25

• 3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes

the word STARTwEND into the first register and returns to the START-cell: START INIT1 ... 26

• 3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes

the word STARTwEND into the first register and returns to the START-cell: START w1 INIT2 ... 27

• 3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes

the word STARTwEND into the first register and returns to the START-cell: START w1

• w2

INIT3

• ...

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• 3. With a probability > 0, the START-cell sends an INIT-signal to the right, which writes

the word STARTwEND into the first register and returns to the START-cell: START w1 w2

• ...
• END

INITn+2

• 29

• 4. When the INIT-signal reaches again the START-cell, a HEAD-signal is started in the

START-cell simulating the head of the LBA A: corresponds to START s0 w1 w2

• ...

wn END 30

• 4. When the INIT-signal reaches again the START-cell, a HEAD-signal is started in the

START-cell simulating the head of the LBA A: corresponds to START v1 v2 s

• ...

vn END So from each configuration from which no configuration containing an X is reachable, a configuration containing the word STARTwEND and a HEAD-signal in the START-cell can be reached. 31

• 5. If the HEAD-signal goes into an accepting state, the cell containing the signal changes to

X: START v1 v2 f+

• ...

vn END 32

• 5. If the HEAD-signal goes into an accepting state, the cell containing the signal changes to

X: START v1

• X ...

vn END So if there is a configuration from which no configuration containing an X is reachable, the HEAD-signal cannot go into an accepting state. If there is none, then the only way to reach a configuration containing X from a configuration containing exactly one START-cell and no signals, is by the HEAD-signal going into an accepting state. 33

It follows that there can only be configurations from which no configuration containing X can be reached, iff the HEAD-signal cannot go into an accepting state ⇔ P is ergodic iff w can be accepted by the LBA A, which proves Theorem 2. 34

Proof of Theorem 3: With a similar construction a Turing machine T with input w can be simulated, so that a PCA with odd length m is ergodic iff T stops for input w on a tape of length m − 2; therefore it is undecidable whether there exists m such that the PCA with length m and given transition rule is ergodic, which proves Theorem 3. 35

• Introduction
• PCAs with 2 states, M1-Neighbourhood
• PCAs with k states, M1-Neighbourhood
• PCAs with 2 states, Mm-Neighbourhood
• Conclusion

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Simulation of a PCA with k states by a PCA P’ with 3 states: P has states {0, ..., k − 1}, P’ has states {0, 1, 2} i → 20i1{0, 1}k−2−i if i < k − 1, k − 1 → 20k−1; e.g. 20011 and 20010 are both coding the state 2 in P. Section of the configuration each cell has to know: ... 200101 200010 211101... 37

If in P a cell changes with probability pj to j, then the corresponding block shall change with the same probability to a block coding the state j. Let ˜ pi be the probability for the ith cell in the block to go into state 1; if the ˜ pi solve the following equations (and solutions always exist), then P’ shows the wanted behaviour: j

i=1(1 − ˜

pi) · ˜ pj+1 = pj for j < k − 1 k−1

i=1 (1 − ˜

pi) = pk−1 Then P’ simulates P. 38

Simulation of a PCA P with k states by a PCA ˜ P with 2 states: P → P’ → ˜ P P’ → ˜ P: 0 → 001, 1 → 011, 2 → 111 Section of the configuration each cell has to know: ... |111 001 011| |111 011 001| |111 001 001|... If a cell in P’ changes with probability p to state 1, then the middle cell of the corresponding triple changes with probability p to 1; all other cells keep their state. ˜ P simulates P’ and therefore P. 39

A configuration of ˜ P is well formed, if it codes a configuration of P. Problem: Even if P is ergodic ˜ P is not ergodic, since the set of all well formed configurations is closed. If a configuration is not well formed, then there are cells which can detect it; all these cells shall change to state 0 with probability 0 < p < 1. ⇒ State 0 can spread over the whole PCA; from 0n every configuration can be reached, and 0n is aperiodic. 40

If there is a state X in P which can change to every state and spread, then the transition rule of ˜ P can be changed by letting all cells in a block coding X change to 0. ⇒ The following configuration is no longer well formed. So it follows: P ergodic ⇔ From each configuration a configuration containing X is reachable ⇔ From every well formed configuration in ˜ P a not well formed configuration is reachable ⇔ From every configuration a not well formed configuration is reachable ⇔ From each configuration 0n is reachable ⇔ ˜ P is ergodic. 41

Corollaries:

• 1. It is undecidable for a given transition rule δ, whether there exists n

such that the PCA (with 2 states and Mm-Neighbourhood) with transition rule δ and length m

2 · n is ergodic.

• 2. The problem to decide whether a given PCA (with 2 states and Mm-Neighbourhood)

is ergodic is PSPACE-complete in (n + m) 42

Conclusion: 2 states, M1-Neighbourhood: Transition rules can be classified as homogenous, inhomogenous, structure-invariant and ergodic. k states, M1-Neighbourhood:

• The problem to decide whether a PCA is ergodic is PSPACE-complete.
• The problem to determine for a given transition rule whether there is a n such that the

corresponding PCA is ergodic is undecidable. 43

2 states, Mm-Neighbourhood:

• PCAs with k states can be simulated.
• The problem to decide whether a PCA is ergodic is PSPACE-complete.
• The problem to determine for a given transition rule whether there is a n such that the

corresponding PCA with length m

2 · n is ergodic is undecidable.

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• Thank you for your attention!

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